- #1

FritoTaco

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## Homework Statement

If a ball is thrown into the air with a velocity of 48 ft/s, its height in feet

*t*seconds later is given by

*y*= 48

*t*− 16

*t*2.

(a) Find the average velocity for the time period beginning when

*t*= 2 and lasting for each of the following.

(i) 0.5 seconds

(ii) 0.1 seconds

(iii) 0.05 seconds

(iv) 0.01 seconds

(b) Estimate the instantaneous velocity when

*t*= 2.

## Homework Equations

[itex]Average Velocity =\dfrac{\triangle y}{\triangle t}[/itex] (Change in position over change in time)

[itex]m=\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}[/itex]

## The Attempt at a Solution

Set: [itex]t = 2[/itex] and [itex]t = 2 + h[/itex] (let h be the time difference) [itex]\ne 0[/itex]

[itex]t = 2[/itex] is [itex]y_{1},x_{1}[/itex] and [itex]t = 2 + h[/itex] is [itex]y_{2},x_{2}[/itex]

Average Velocity = [itex]\dfrac{y(2+h)-y(2)}{(2+h)-2}[/itex]

Now with [itex]y=48t-16t^2[/itex], we will replace [itex]t[/itex] with [itex]2[/itex] and [itex]2 + h[/itex].

[itex]\dfrac{48(2+h)-16(2+h)^2-[48(2)-16(2)^2)]}{h}[/itex]

[itex]\dfrac{96+48h-16(4+4h+h^2)-96+64}{h}[/itex]

[itex]\dfrac{48h-64-64h-16h^2+64}{h}[/itex]

[itex]\dfrac{-16h^2-16h}{h}[/itex]

[itex]\dfrac{h(-16h-16)}{h}[/itex]

Average Velocity =[itex]-16h-16[/itex]

Then I plug in (i) 0.5 into the equation like this: [itex]-16(0.5)-16=24[/itex]

I do this for the rest of them and get correct answers according to the online "WebAssign" website. Now I'm stuck on (b) "Estimate the instantaneous velocity when

*t*= 2." You can look at my picture, I tried plugging in values near 2 like 2.1 and 2.01, etc, into [itex]y=48t-16t^2[/itex] What am I doing wrong? So as I get closer to 2 I get closer to the slope of 31.999 which then you make an educated guess (by rounding) and get 32. But this is wrong. I don't know why? Also, I heard there is another method for finding average velocity, my book doesn't have it--anyone know it? Much appreciation.